Variable separable type differential equations Questions in English

(A) Given the differential equation: $3f^2(x) f'(x) = 2x$.
Integrating both sides with respect to $x$:
$\int 3f^2(x) f'(x) dx = \int 2x dx$.
Using the substitution $u = f(x)$,$du = f'(x) dx$,we get:
$\int 3u^2 du = x^2 + C$.
$u^3 = x^2 + C$,which means $f^3(x) = x^2 + C$.
Given $f(2) = 1$,substitute $x = 2$ and $f(2) = 1$:
$1^3 = 2^2 + C
1 = 4 + C
C = -3$.
So,the function is $f^3(x) = x^2 - 3$.
To find $f(3)$,substitute $x = 3$:
$f^3(3) = 3^2 - 3 = 9 - 3 = 6$.
Therefore,$f(3) = \sqrt[3]{6}$.

(B) Given the slope of the tangent is $\frac{dy}{dx} = \cos(x + y) + \sin(x + y)$.
Let $u = x + y$,then $\frac{du}{dx} = 1 + \frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{du}{dx} - 1$.
Substituting this into the equation: $\frac{du}{dx} - 1 = \cos u + \sin u$.
$\frac{du}{dx} = 1 + \cos u + \sin u = 2 \cos^2(\frac{u}{2}) + 2 \sin(\frac{u}{2}) \cos(\frac{u}{2}) = 2 \cos^2(\frac{u}{2}) [1 + \tan(\frac{u}{2})]$.
Separating variables: $\int \frac{\sec^2(\frac{u}{2})}{2(1 + \tan(\frac{u}{2}))} du = \int dx$.
Let $t = \tan(\frac{u}{2})$,then $dt = \frac{1}{2} \sec^2(\frac{u}{2}) du$.
The integral becomes $\int \frac{dt}{1 + t} = \int dx$,which gives $\ln(1 + t) = x + C$.
Since the curve passes through the origin $(0, 0)$,$u = x + y = 0$,so $t = \tan(0) = 0$.
Substituting $x = 0, t = 0$ into $\ln(1 + t) = x + C$ gives $C = 0$.
Thus,$\ln(1 + t) = x$,which implies $1 + t = e^x$,or $t = e^x - 1$.
Substituting back $t = \tan(\frac{x + y}{2})$,we get $\tan(\frac{x + y}{2}) = e^x - 1$.
Therefore,$\frac{x + y}{2} = \tan^{-1}(e^x - 1)$,which simplifies to $y = 2 \tan^{-1}(e^x - 1) - x$.

(A) Given the differential equation: $y \ln y + x \frac{dy}{dx} = 0$.
Rearranging the terms to separate the variables: $\frac{dx}{x} + \frac{dy}{y \ln y} = 0$.
Integrating both sides: $\int \frac{dx}{x} + \int \frac{dy}{y \ln y} = C$.
Let $u = \ln y$,then $du = \frac{1}{y} dy$. The integral becomes: $\ln |x| + \ln |\ln y| = C$.
This simplifies to: $\ln |x \ln y| = C$,or $x \ln y = k$ where $k = e^C$.
Using the initial condition $y(1) = e$: $1 \cdot \ln(e) = k \Rightarrow 1 \cdot 1 = k \Rightarrow k = 1$.
Thus,the solution is $x \ln y = 1$.

(A) Given the differential equation: $2x^2y \frac{dy}{dx} = \tan(x^2y^2) - 2xy^2$.
Rearranging the terms,we get: $2x^2y \frac{dy}{dx} + 2xy^2 = \tan(x^2y^2)$.
Notice that the left side is the derivative of $x^2y^2$ with respect to $x$: $\frac{d}{dx}(x^2y^2) = 2x^2y \frac{dy}{dx} + 2xy^2$.
Let $z = x^2y^2$. Then the equation becomes $\frac{dz}{dx} = \tan z$.
Separating the variables: $\int \cot z \, dz = \int dx$.
Integrating both sides: $\ln(\sin z) = x + C$.
Substituting $z = x^2y^2$: $\ln(\sin(x^2y^2)) = x + C$.
Given $y(1) = \sqrt{\frac{\pi}{2}}$,at $x=1$,$z = (1)^2 \cdot (\sqrt{\frac{\pi}{2}})^2 = \frac{\pi}{2}$.
Substituting these values: $\ln(\sin(\frac{\pi}{2})) = 1 + C \Rightarrow \ln(1) = 1 + C \Rightarrow 0 = 1 + C \Rightarrow C = -1$.
Thus,$\ln(\sin(x^2y^2)) = x - 1$.
Taking the exponential of both sides: $\sin(x^2y^2) = e^{x-1}$.

(A) Given the differential equation $\frac{dy}{dx} = \frac{1 - 2(2x + y)}{1 + (2x + y)}$.
Let $v = 2x + y$. Then $\frac{dv}{dx} = 2 + \frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{dv}{dx} - 2$.
Substituting these into the equation: $\frac{dv}{dx} - 2 = \frac{1 - 2v}{1 + v}$.
$\frac{dv}{dx} = \frac{1 - 2v}{1 + v} + 2 = \frac{1 - 2v + 2 + 2v}{1 + v} = \frac{3}{1 + v}$.
Separating variables: $(1 + v) dv = 3 dx$.
Integrating both sides: $\int (1 + v) dv = \int 3 dx$.
$v + \frac{v^2}{2} = 3x + c_1 \Rightarrow 2v + v^2 = 6x + 2c_1$.
Substitute $v = 2x + y$ back: $2(2x + y) + (2x + y)^2 = 6x + C$.
$4x + 2y + 4x^2 + 4xy + y^2 = 6x + C$.
$4x^2 + 4xy + y^2 - 2x + 2y + C = 0$.

(C) Given the differential equation $\frac{dy}{dx} = \frac{y - x}{y - x - 1}$.
Let $y - x = t$,then $\frac{dy}{dx} - 1 = \frac{dt}{dx}$,so $\frac{dy}{dx} = \frac{dt}{dx} + 1$.
Substituting this into the equation: $\frac{dt}{dx} + 1 = \frac{t}{t - 1}$.
$\frac{dt}{dx} = \frac{t}{t - 1} - 1 = \frac{t - (t - 1)}{t - 1} = \frac{1}{t - 1}$.
Separating variables: $(t - 1) dt = dx$.
Integrating both sides: $\int (t - 1) dt = \int dx \Rightarrow \frac{t^2}{2} - t = x + C$.
Substituting $t = y - x$: $\frac{(y - x)^2}{2} - (y - x) = x + C$.
$(y - x)^2 - 2(y - x) = 2x + 2C \Rightarrow (y - x)^2 - 2y + 2x = 2x + 2C \Rightarrow (y - x)^2 - 2y = K$.
Given $y(-5) = -5$,so $(-5 - (-5))^2 - 2(-5) = K \Rightarrow 0 + 10 = K \Rightarrow K = 10$.
The equation is $(y - x)^2 - 2y = 10$,which is $y^2 - 2xy + x^2 - 2y - 10 = 0$.
The discriminant $h^2 - ab = (-1)^2 - (1)(1) = 1 - 1 = 0$. Since the discriminant is $0$,the equation represents a parabola.

(D) Given the integral equation: $\int\limits_0^x {t\,y(t)\,dt} = x^2y(x)$.
Differentiating both sides with respect to $x$ using the Leibniz rule:
$x\,y(x) = \frac{d}{dx}(x^2y(x))$
$x\,y(x) = x^2y'(x) + 2x\,y(x)$.
Rearranging the terms:
$x^2y'(x) + x\,y(x) = 0$.
Dividing by $x$ (since $x > 0$):
$x\,y'(x) + y(x) = 0$.
This is a separable differential equation:
$x\frac{dy}{dx} = -y$
$\frac{dy}{y} = -\frac{dx}{x}$.
Integrating both sides:
$\ln|y| = -\ln|x| + C$
$\ln|y| + \ln|x| = C$
$\ln|xy| = C$
$xy = k$,where $k = e^C$.
Since the curve passes through $(2, 3)$:
$(2)(3) = k \implies k = 6$.
Thus,the equation of the curve is $xy = 6$.

(D) Given the differential equation: $[f'(x)]^2 + 4f(x)f'(x) + f^2(x) = 0$.
Let $y = f(x)$,then $f'(x) = \frac{dy}{dx}$. The equation becomes: $(\frac{dy}{dx})^2 + 4y(\frac{dy}{dx}) + y^2 = 0$.
This is a quadratic equation in $\frac{dy}{dx}$. Using the quadratic formula $\frac{dy}{dx} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$\frac{dy}{dx} = \frac{-4y \pm \sqrt{(4y)^2 - 4(1)(y^2)}}{2(1)} = \frac{-4y \pm \sqrt{16y^2 - 4y^2}}{2} = \frac{-4y \pm \sqrt{12y^2}}{2} = \frac{-4y \pm 2\sqrt{3}y}{2} = (-2 \pm \sqrt{3})y$.
Case $1$: $\frac{dy}{dx} = (-2 + \sqrt{3})y$. Separating variables: $\int \frac{dy}{y} = \int (-2 + \sqrt{3}) dx$.
Integrating both sides: $\ln|y| = (-2 + \sqrt{3})x + k_1 \implies y = c_1 e^{(\sqrt{3} - 2)x}$.
Case $2$: $\frac{dy}{dx} = (-2 - \sqrt{3})y$. Separating variables: $\int \frac{dy}{y} = \int (-2 - \sqrt{3}) dx$.
Integrating both sides: $\ln|y| = -(2 + \sqrt{3})x + k_2 \implies y = c_2 e^{-(2 + \sqrt{3})x}$.
Thus,both solutions are valid. Therefore,the correct option is $(D)$.

(C) The given equation is $x = 1 + xy\frac{dy}{dx} + \frac{(xy)^2}{2!}\left(\frac{dy}{dx}\right)^2 + \dots$
This is the Taylor series expansion of the exponential function $e^u$,where $u = xy\frac{dy}{dx}$.
Thus,the equation can be written as $x = e^{xy\frac{dy}{dx}}$.
Taking the natural logarithm on both sides,we get $\log_e x = xy\frac{dy}{dx}$.
Rearranging the terms to separate the variables,we have $y \, dy = \frac{\log_e x}{x} \, dx$.
Integrating both sides,$\int y \, dy = \int \frac{\log_e x}{x} \, dx$.
Let $t = \log_e x$,then $dt = \frac{1}{x} \, dx$.
So,$\frac{y^2}{2} = \frac{(\log_e x)^2}{2} + C$.
Multiplying by $2$,we get $y^2 = (\log_e x)^2 + 2C$.
Therefore,$y = \pm \sqrt{(\log_e x)^2 + 2C}$.

(A) Given the differential equation: $(2 + \cos x)\frac{dz}{dx} + (\sin x)z = \sin x$.
Rearranging the terms: $(2 + \cos x)\frac{dz}{dx} = \sin x - (\sin x)z = -\sin x(z - 1)$.
Separating the variables: $\frac{dz}{z - 1} = \frac{-\sin x}{2 + \cos x} dx$.
Integrating both sides: $\int \frac{dz}{z - 1} = \int \frac{-\sin x}{2 + \cos x} dx$.
Let $u = 2 + \cos x$,then $du = -\sin x dx$.
So,$\ln|z - 1| = \ln|2 + \cos x| + C$.
Using the condition $z(\frac{\pi}{2}) = 3$: $\ln|3 - 1| = \ln|2 + \cos(\frac{\pi}{2})| + C \implies \ln 2 = \ln 2 + C \implies C = 0$.
Thus,$z - 1 = 2 + \cos x \implies z = 3 + \cos x$.
Now,find $z(\frac{\pi}{3})$: $z(\frac{\pi}{3}) = 3 + \cos(\frac{\pi}{3}) = 3 + \frac{1}{2} = \frac{7}{2}$.

(B) Given the differential equation: $x dy + y dx - \sqrt{1 - x^2 y^2} dx = 0$.
Rearranging the terms,we get: $x dy + y dx = \sqrt{1 - x^2 y^2} dx$.
We know that $d(xy) = x dy + y dx$.
Substituting this into the equation: $d(xy) = \sqrt{1 - (xy)^2} dx$.
Dividing both sides by $\sqrt{1 - (xy)^2}$,we get: $\frac{d(xy)}{\sqrt{1 - (xy)^2}} = dx$.
Integrating both sides: $\int \frac{d(xy)}{\sqrt{1 - (xy)^2}} = \int dx$.
This results in: $\sin^{-1}(xy) = x + c$.
Taking the sine of both sides: $xy = \sin(x + c)$.

(B) Given equation: $\left( {1 + {e^{2y}}} \right){e^{{{\tan }^{ - 1}}x}}dx = \left( {1 + {x^2}} \right)\left( {{e^y} + {{\left( {{e^y} - 1} \right)}^2}} \right)dy$
Rearranging terms: $\frac{{{e^{{{\tan }^{ - 1}}x}}}}{{1 + {x^2}}}dx = \frac{{{e^y} + {e^{2y}} - 2{e^y} + 1}}{{1 + {e^{2y}}}}dy = \frac{{{e^{2y}} - {e^y} + 1}}{{1 + {e^{2y}}}}dy$
Integrating both sides: $\int \frac{{{e^{{{\tan }^{ - 1}}x}}}}{{1 + {x^2}}}dx = \int \frac{{{e^{2y}} - {e^y} + 1}}{{1 + {e^{2y}}}}dy$
Let $u = \tan^{-1}x$,then $du = \frac{1}{1+x^2}dx$. The $LHS$ becomes $\int e^u du = e^u = e^{\tan^{-1}x}$.
For the $RHS$: $\int \frac{e^{2y}+1-e^y}{1+e^{2y}} dy = \int (1 - \frac{e^y}{1+e^{2y}}) dy = y - \int \frac{e^y}{1+(e^y)^2} dy$.
Let $v = e^y$,then $dv = e^y dy$. The integral becomes $y - \tan^{-1}(e^y) + C$.
Equating both sides: $e^{\tan^{-1}x} = y - \tan^{-1}(e^y) + C$.
Rearranging: $\tan^{-1}(e^y) = y - e^{\tan^{-1}x} + C$.
Taking tangent on both sides: $e^y = \tan(y - e^{\tan^{-1}x} + C)$.
Taking natural log: $y = \ln(\tan(y - e^{\tan^{-1}x} + C))$.

(A) Given the differential equation: $e^{2y}(1 + \ln x)dx + \csc y(2 + \cot y)dy = 0$.
Rearranging the terms,we get: $\csc y(2 + \cot y)dy = -e^{2y}(1 + \ln x)dx$.
Dividing by $e^{2y}$,we have: $e^{-2y}(\csc y(2 + \cot y))dy = -(1 + \ln x)dx$.
Integrating both sides: $\int e^{-2y}(2\csc y + \csc y \cot y)dy = -\int (1 + \ln x)dx$.
Let $f(y) = e^{-2y} \csc y$. Then $f'(y) = e^{-2y}(-\csc y \cot y) + (-2)e^{-2y} \csc y = -e^{-2y}(\csc y \cot y + 2 \csc y)$.
Thus,the integral of the left side is $-e^{-2y} \csc y$.
The integral of the right side is $-\int (1 + \ln x)dx = -(x \ln x) + C$.
So,$-e^{-2y} \csc y = -x \ln x + C$,which simplifies to $x \ln x + C = e^{-2y} \csc y = \frac{e^{-2y}}{\sin y}$.
Using the condition $y(1) = \frac{\pi}{2}$,we substitute $x=1$ and $y=\frac{\pi}{2}$:
$1 \cdot \ln(1) + C = \frac{e^{-2(\pi/2)}}{\sin(\pi/2)} \Rightarrow 0 + C = \frac{e^{-\pi}}{1} \Rightarrow C = e^{-\pi}$.
Therefore,the solution is $x \ln x + e^{-\pi} = \frac{e^{-2y}}{\sin y}$.

(D) Given the differential equation: $\frac{dy}{dx} = \frac{2x}{3y}$.
Separating the variables,we get: $3y \, dy = 2x \, dx$.
Integrating both sides: $\int 3y \, dy = \int 2x \, dx$.
This gives: $\frac{3y^2}{2} = x^2 + C_1$,or $3y^2 = 2x^2 + C$.
Rearranging the terms: $2x^2 - 3y^2 = -C$,which is of the form $Ax^2 - By^2 = K$.
Since the curve passes through $(\sqrt{2}, 1)$,substitute these values: $2(\sqrt{2})^2 - 3(1)^2 = K \implies 2(2) - 3 = K \implies K = 1$.
The equation of the curve is $2x^2 - 3y^2 = 1$.
This equation represents a hyperbola.

(A) Given the differential equation: $x \frac{dy}{dx} + y = x \frac{f(xy)}{f'(xy)}$.
We know that $x dy + y dx = d(xy)$.
Rearranging the given equation: $x dy + y dx = x \frac{f(xy)}{f'(xy)} dx$.
This is not quite right,let's rewrite: $\frac{dy}{dx} + \frac{y}{x} = \frac{f(xy)}{f'(xy)}$.
Multiply by $x$: $x \frac{dy}{dx} + y = x \frac{f(xy)}{f'(xy)}$.
Note that $x \frac{dy}{dx} + y = \frac{d}{dx}(xy)$.
So,$\frac{d(xy)}{dx} = x \frac{f(xy)}{f'(xy)}$.
Rearranging the terms: $\frac{f'(xy)}{f(xy)} d(xy) = x dx$.
Integrating both sides: $\int \frac{f'(xy)}{f(xy)} d(xy) = \int x dx$.
$\ln|f(xy)| = \frac{x^2}{2} + C$.
Taking the exponential of both sides: $f(xy) = e^{\frac{x^2}{2} + C} = k e^{x^2/2}$,where $k = e^C$.

(A) The given differential equation is $xdx + ydy = \frac{xdy - ydx}{x^2 + y^2}$.
We know that $d(x^2 + y^2) = 2xdx + 2ydy$,which implies $xdx + ydy = \frac{1}{2} d(x^2 + y^2)$.
Also,we know that $d(\tan^{-1}(y/x)) = \frac{1}{1 + (y/x)^2} \cdot d(y/x) = \frac{x^2}{x^2 + y^2} \cdot \frac{xdy - ydx}{x^2} = \frac{xdy - ydx}{x^2 + y^2}$.
Substituting these into the given equation,we get:
$\frac{1}{2} d(x^2 + y^2) = d(\tan^{-1}(y/x))$.
Integrating both sides,we get:
$\frac{1}{2}(x^2 + y^2) = \tan^{-1}(y/x) + c$.

(B) Given the differential equation $\frac{d}{dy} \left( \int_x^y dt \right) = x$.
Using the Fundamental Theorem of Calculus,$\frac{d}{dy} \int_x^y dt = 1 - \frac{dx}{dy}$.
Substituting this into the equation: $1 - \frac{dx}{dy} = x$.
Rearranging the terms: $\frac{dx}{dy} = 1 - x$.
Separating the variables: $\frac{dx}{1 - x} = dy$.
Integrating both sides: $\int \frac{dx}{1 - x} = \int dy$.
$-\ln|1 - x| = y + C_1$,which can be written as $y = -\ln|1 - x| + C$.

(B) Given the differential equation: $\sec^2 x \tan y \, dx + \sec^2 y \tan x \, dy = 0$
Rearranging the terms to separate the variables:
$\sec^2 x \tan y \, dx = -\sec^2 y \tan x \, dy$
Dividing both sides by $\tan x \tan y$:
$\frac{\sec^2 x}{\tan x} \, dx = -\frac{\sec^2 y}{\tan y} \, dy$
Integrating both sides:
$\int \frac{\sec^2 x}{\tan x} \, dx = -\int \frac{\sec^2 y}{\tan y} \, dy$
Let $u = \tan x$,then $du = \sec^2 x \, dx$. Let $v = \tan y$,then $dv = \sec^2 y \, dy$:
$\ln |\tan x| = -\ln |\tan y| + C$
$\ln |\tan x| + \ln |\tan y| = C$
$\ln |\tan x \tan y| = C$
$\tan x \tan y = e^C = K$
Using the initial condition $y(\frac{\pi}{4}) = \frac{\pi}{3}$:
$\tan(\frac{\pi}{4}) \tan(\frac{\pi}{3}) = K$
$1 \times \sqrt{3} = K \implies K = \sqrt{3}$
Thus,the solution is $\tan x \tan y = \sqrt{3}$.

(C) Given the differential equation: $\frac{dy}{dx} = \frac{(1+x)y}{(y-1)x}$.
Separating the variables,we get:
$\frac{y-1}{y} dy = \frac{1+x}{x} dx$
Integrating both sides:
$\int (1 - \frac{1}{y}) dy = \int (1 + \frac{1}{x}) dx$
$y - \log|y| = x + \log|x| + c$
Rearranging the terms:
$y - x - c = \log|x| + \log|y|$
$y - x - c = \log|xy|$
$\log|xy| + x - y = -c$
Since $-c$ is also an arbitrary constant,we can write it as $C$:
$\log|xy| + x - y = C$.

(D) Given the differential equation: $\frac{dy}{dx} = \frac{y(x+1)}{x(y+1)}$
Separating the variables,we get:
$\frac{y+1}{y} dy = \frac{x+1}{x} dx$
Integrating both sides:
$\int (1 + \frac{1}{y}) dy = \int (1 + \frac{1}{x}) dx$
$y + \ln|y| = x + \ln|x| + C$
Rearranging the terms:
$y - x = \ln|x| - \ln|y| + C$
$y - x = \ln|\frac{x}{y}| + C$
Since this result is not present in the given options,the correct choice is $D$.

(A) The slope of the curve is given by $\frac{dy}{dx} = 1 - \frac{1}{x^2}$.
Integrating both sides with respect to $x$:
$\int dy = \int \left( 1 - \frac{1}{x^2} \right) dx$
$y = x + \frac{1}{x} + C$
Since the curve passes through the point $\left( 2, \frac{7}{2} \right)$,we substitute these values to find $C$:
$\frac{7}{2} = 2 + \frac{1}{2} + C$
$\frac{7}{2} = \frac{5}{2} + C$
$C = 1$
Thus,the equation of the curve is $y = x + \frac{1}{x} + 1$.
To find the ordinate when the abscissa is $x = -2$:
$y = -2 + \frac{1}{-2} + 1$
$y = -2 - 0.5 + 1 = -1.5 = -\frac{3}{2}$.

(C) The given differential equation is $\frac{(2 + \sin x)}{(1 + y)} \frac{dy}{dx} = \cos x.$
Separating the variables,we get $\frac{dy}{1 + y} = \frac{\cos x}{2 + \sin x} dx.$
Integrating both sides,we have $\int \frac{dy}{1 + y} = \int \frac{\cos x}{2 + \sin x} dx.$
This yields $\ln(1 + y) = \ln(2 + \sin x) + \ln C.$
Exponentiating both sides,we get $1 + y = C(2 + \sin x).$
Given $y(0) = 2,$ we substitute $x = 0$ and $y = 2$ to find $C$: $1 + 2 = C(2 + \sin 0) \Rightarrow 3 = 2C \Rightarrow C = \frac{3}{2}.$
Now,we find $y\left( \frac{\pi}{2} \right)$ by substituting $x = \frac{\pi}{2}$ and $C = \frac{3}{2}$ into the equation $1 + y = \frac{3}{2}(2 + \sin x).$
$1 + y\left( \frac{\pi}{2} \right) = \frac{3}{2}(2 + \sin \frac{\pi}{2}) = \frac{3}{2}(2 + 1) = \frac{3}{2}(3) = \frac{9}{2}.$
Therefore,$y\left( \frac{\pi}{2} \right) = \frac{9}{2} - 1 = \frac{7}{2}.$

(B) Let $u = x - y$. Then $\frac{du}{dx} = 1 - \frac{dy}{dx}$.
Substituting $\frac{dy}{dx} = u^2$,we get $1 - \frac{du}{dx} = u^2$,which implies $\frac{du}{dx} = 1 - u^2$.
Separating the variables,we have $\frac{du}{1 - u^2} = dx$.
Integrating both sides,$\int \frac{du}{1 - u^2} = \int dx$,which gives $\frac{1}{2} \log_e \left| \frac{1 + u}{1 - u} \right| = x + C$.
Substituting $u = x - y$,we get $\frac{1}{2} \log_e \left| \frac{1 + x - y}{1 - x + y} \right| = x + C$.
Using the condition $y(1) = 1$,we have $x = 1, y = 1$,so $u = 1 - 1 = 0$.
$\frac{1}{2} \log_e \left| \frac{1 + 0}{1 - 0} \right| = 1 + C \Rightarrow 0 = 1 + C \Rightarrow C = -1$.
Thus,$\frac{1}{2} \log_e \left| \frac{1 + x - y}{1 - x + y} \right| = x - 1$,which simplifies to $\log_e \left| \frac{1 + x - y}{1 - x + y} \right| = 2(x - 1)$.
Multiplying by $-1$ on both sides,we get $- \log_e \left| \frac{1 + x - y}{1 - x + y} \right| = -2(x - 1)$,which is equivalent to $- \log_e \left| \frac{1 - x + y}{1 + x - y} \right| = 2(x - 1)$.

(B) Given the differential equation: $\sqrt{1-x^{2}} \frac{dy}{dx} + \sqrt{1-y^{2}} = 0$.
Separating the variables,we get: $\frac{dy}{\sqrt{1-y^{2}}} = -\frac{dx}{\sqrt{1-x^{2}}}$.
Integrating both sides: $\int \frac{dy}{\sqrt{1-y^{2}}} = -\int \frac{dx}{\sqrt{1-x^{2}}}$.
This gives: $\sin^{-1} y = -\sin^{-1} x + C$,or $\sin^{-1} x + \sin^{-1} y = C$.
Given $y\left(\frac{1}{2}\right) = \frac{\sqrt{3}}{2}$,substitute $x = \frac{1}{2}$ and $y = \frac{\sqrt{3}}{2}$:
$\sin^{-1}\left(\frac{1}{2}\right) + \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = C$.
$\frac{\pi}{6} + \frac{\pi}{3} = C \implies C = \frac{\pi}{2}$.
So,the equation is $\sin^{-1} x + \sin^{-1} y = \frac{\pi}{2}$.
Using the identity $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$,we have $\sin^{-1} y = \cos^{-1} x$,which implies $y = \sin(\cos^{-1} x) = \sqrt{1-x^2}$.
Now,to find $y\left(-\frac{1}{\sqrt{2}}\right)$,substitute $x = -\frac{1}{\sqrt{2}}$:
$y = \sqrt{1 - \left(-\frac{1}{\sqrt{2}}\right)^2} = \sqrt{1 - \frac{1}{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.

(N/A) Given the differential equation:
$\frac{dy}{dx} = \frac{x+1}{2-y}$
Separating the variables,we get:
$(2-y) dy = (x+1) dx$
Integrating both sides:
$\int (2-y) dy = \int (x+1) dx$
Performing the integration:
$2y - \frac{y^2}{2} = \frac{x^2}{2} + x + C_1$
Multiplying the entire equation by $2$:
$4y - y^2 = x^2 + 2x + 2C_1$
Rearranging the terms to form the general solution:
$x^2 + y^2 + 2x - 4y + C = 0$,where $C = 2C_1$.

(A) Given the differential equation: $\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$.
Separating the variables,we get:
$\frac{dy}{1+y^2} = \frac{dx}{1+x^2}$.
Integrating both sides:
$\int \frac{dy}{1+y^2} = \int \frac{dx}{1+x^2}$.
Using the standard integral formula $\int \frac{du}{1+u^2} = \tan^{-1} u + C$,we obtain:
$\tan^{-1} y = \tan^{-1} x + C$,where $C$ is the constant of integration.

(A) Given the differential equation $\frac{dy}{dx} = -4xy^2$.
Separating the variables,we get $\frac{dy}{y^2} = -4x dx$.
Integrating both sides,we have $\int y^{-2} dy = -4 \int x dx$.
This gives $-\frac{1}{y} = -4 \cdot \frac{x^2}{2} + C$,which simplifies to $-\frac{1}{y} = -2x^2 + C$.
Multiplying by $-1$,we get $\frac{1}{y} = 2x^2 - C$.
Given that $y = 1$ when $x = 0$,we substitute these values: $\frac{1}{1} = 2(0)^2 - C$,which implies $1 = -C$,so $C = -1$.
Substituting $C = -1$ back into the equation $\frac{1}{y} = 2x^2 - C$,we get $\frac{1}{y} = 2x^2 - (-1) = 2x^2 + 1$.
Therefore,the particular solution is $y = \frac{1}{2x^2 + 1}$.

(A) The given differential equation is $x dy = (2x^2 + 1) dx$.
Dividing both sides by $x$ (since $x \neq 0$),we get $dy = (2x + \frac{1}{x}) dx$.
Integrating both sides,we have $\int dy = \int (2x + \frac{1}{x}) dx$.
This gives $y = x^2 + \log |x| + C$.
Since the curve passes through the point $(1, 1)$,we substitute $x = 1$ and $y = 1$ into the equation:
$1 = (1)^2 + \log |1| + C$.
$1 = 1 + 0 + C$,which implies $C = 0$.
Substituting $C = 0$ back into the general solution,we get the required equation: $y = x^2 + \log |x|$.

(A) The slope of the tangent to the curve at any point $(x, y)$ is given by $\frac{dy}{dx} = \frac{2x}{y^2}$.
Separating the variables,we get $y^2 dy = 2x dx$.
Integrating both sides,we have $\int y^2 dy = \int 2x dx$.
This gives $\frac{y^3}{3} = x^2 + C$.
Since the curve passes through the point $(-2, 3)$,we substitute $x = -2$ and $y = 3$ into the equation:
$\frac{3^3}{3} = (-2)^2 + C$
$\frac{27}{3} = 4 + C$
$9 = 4 + C$
$C = 5$.
Substituting $C = 5$ back into the equation,we get $\frac{y^3}{3} = x^2 + 5$.

(A) The given differential equation is: $\frac{dy}{dx} = \frac{1-\cos x}{1+\cos x}$.
Using trigonometric identities $1-\cos x = 2 \sin^2 \frac{x}{2}$ and $1+\cos x = 2 \cos^2 \frac{x}{2}$,we get:
$\frac{dy}{dx} = \frac{2 \sin^2 \frac{x}{2}}{2 \cos^2 \frac{x}{2}} = \tan^2 \frac{x}{2}$.
Since $\tan^2 \theta = \sec^2 \theta - 1$,we have:
$\frac{dy}{dx} = \sec^2 \frac{x}{2} - 1$.
Separating the variables,we get:
$dy = (\sec^2 \frac{x}{2} - 1) dx$.
Integrating both sides:
$\int dy = \int (\sec^2 \frac{x}{2} - 1) dx$.
$y = 2 \tan \frac{x}{2} - x + C$.
Thus,the general solution is $y = 2 \tan \frac{x}{2} - x + C$.

(A) The given differential equation is: $\frac{dy}{dx} = \sqrt{4 - y^2}$.
Separating the variables,we get:
$\frac{dy}{\sqrt{4 - y^2}} = dx$.
Integrating both sides with respect to their variables:
$\int \frac{dy}{\sqrt{2^2 - y^2}} = \int dx$.
Using the standard integral formula $\int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}(\frac{u}{a}) + C$,we get:
$\sin^{-1}(\frac{y}{2}) = x + C$.
Taking the sine of both sides:
$\frac{y}{2} = \sin(x + C)$.
Thus,the general solution is:
$y = 2 \sin(x + C)$.

(A) The given differential equation is: $\frac{dy}{dx} + y = 1$ $(y \neq 1)$.
Rearranging the terms,we get:
$\frac{dy}{dx} = 1 - y$
Separating the variables,we get:
$\frac{dy}{1 - y} = dx$
Integrating both sides:
$\int \frac{dy}{1 - y} = \int dx$
$-\log|1 - y| = x + C_1$
$\log|1 - y| = -x - C_1$
Taking the exponential of both sides:
$|1 - y| = e^{-x - C_1} = e^{-C_1} \cdot e^{-x}$
$1 - y = \pm e^{-C_1} e^{-x}$
Let $A = \mp e^{-C_1}$,then:
$1 - y = Ae^{-x}$
$y = 1 - Ae^{-x}$
Since $A$ is an arbitrary constant,we can write the general solution as $y = 1 + Ce^{-x}$.

(A) The given differential equation is: $\sec^{2} x \tan y \, dx + \sec^{2} y \tan x \, dy = 0$.
Dividing both sides by $\tan x \tan y$,we get:
$\frac{\sec^{2} x}{\tan x} \, dx + \frac{\sec^{2} y}{\tan y} \, dy = 0$.
Integrating both sides:
$\int \frac{\sec^{2} x}{\tan x} \, dx + \int \frac{\sec^{2} y}{\tan y} \, dy = C_1$.
Let $u = \tan x$,then $du = \sec^{2} x \, dx$.
Let $v = \tan y$,then $dv = \sec^{2} y \, dy$.
Substituting these into the integral:
$\int \frac{1}{u} \, du + \int \frac{1}{v} \, dv = C_1$.
$\ln|u| + \ln|v| = C_1$.
$\ln|\tan x| + \ln|\tan y| = C_1$.
$\ln|\tan x \tan y| = C_1$.
Taking the exponential of both sides:
$\tan x \tan y = e^{C_1} = C$.

(A) The given differential equation is: $(e^{x}+e^{-x}) dy = (e^{x}-e^{-x}) dx$.
Separating the variables,we get: $dy = \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} dx$.
Integrating both sides: $\int dy = \int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} dx + C$.
Let $t = e^{x}+e^{-x}$. Then $dt = (e^{x}-e^{-x}) dx$.
Substituting these into the integral: $y = \int \frac{1}{t} dt + C$.
Integrating,we get: $y = \log|t| + C$.
Substituting back $t = e^{x}+e^{-x}$,the general solution is: $y = \log(e^{x}+e^{-x}) + C$.

(A) The given differential equation is: $\frac{dy}{dx} = (1+x^2)(1+y^2)$
Separating the variables,we get: $\frac{dy}{1+y^2} = (1+x^2) dx$
Integrating both sides: $\int \frac{dy}{1+y^2} = \int (1+x^2) dx$
We know that $\int \frac{dy}{1+y^2} = \tan^{-1} y$ and $\int (1+x^2) dx = x + \frac{x^3}{3} + C$
Thus,the general solution is: $\tan^{-1} y = x + \frac{x^3}{3} + C$

(A) The given differential equation is: $y \log y \, dx - x \, dy = 0$.
Rearranging the terms,we get: $y \log y \, dx = x \, dy$.
Separating the variables,we have: $\frac{dy}{y \log y} = \frac{dx}{x}$.
Integrating both sides: $\int \frac{dy}{y \log y} = \int \frac{dx}{x}$.
Let $t = \log y$,then $dt = \frac{1}{y} \, dy$.
Substituting this into the integral: $\int \frac{dt}{t} = \int \frac{dx}{x}$.
Integrating,we get: $\log |t| = \log |x| + \log |C|$.
$\log |\log y| = \log |Cx|$.
Taking the exponential of both sides: $\log y = Cx$.
Thus,the general solution is: $y = e^{Cx}$.

(A) The given differential equation is: $x^{5} \frac{dy}{dx} = -y^{5}$
Separating the variables,we get: $\frac{dy}{y^{5}} = -\frac{dx}{x^{5}}$
Rearranging the terms: $\frac{dx}{x^{5}} + \frac{dy}{y^{5}} = 0$
Integrating both sides: $\int x^{-5} dx + \int y^{-5} dy = k$ (where $k$ is an arbitrary constant)
Applying the power rule for integration $\int x^{n} dx = \frac{x^{n+1}}{n+1}$: $\frac{x^{-4}}{-4} + \frac{y^{-4}}{-4} = k$
Multiplying by $-4$: $x^{-4} + y^{-4} = -4k$
Let $C = -4k$,then the general solution is: $x^{-4} + y^{-4} = C$

(A) The given differential equation is: $\frac{dy}{dx} = \sin^{-1} x$
Integrating both sides with respect to $x$:
$y = \int \sin^{-1} x \, dx$
Using integration by parts,let $u = \sin^{-1} x$ and $dv = dx$. Then $du = \frac{1}{\sqrt{1-x^2}} dx$ and $v = x$.
$y = x \sin^{-1} x - \int x \cdot \frac{1}{\sqrt{1-x^2}} dx$
Let $t = 1 - x^2$,then $dt = -2x \, dx$,which implies $x \, dx = -\frac{1}{2} dt$.
$y = x \sin^{-1} x - \int \frac{-1/2}{\sqrt{t}} dt$
$y = x \sin^{-1} x + \frac{1}{2} \int t^{-1/2} dt$
$y = x \sin^{-1} x + \frac{1}{2} \cdot \frac{t^{1/2}}{1/2} + C$
$y = x \sin^{-1} x + \sqrt{t} + C$
$y = x \sin^{-1} x + \sqrt{1 - x^2} + C$

(A) The given differential equation is: $e^{x} \tan y \, dx + (1 - e^{x}) \sec^{2} y \, dy = 0$.
Rearranging the terms,we get: $(1 - e^{x}) \sec^{2} y \, dy = -e^{x} \tan y \, dx$.
Separating the variables,we get: $\frac{\sec^{2} y}{\tan y} \, dy = \frac{-e^{x}}{1 - e^{x}} \, dx$.
Integrating both sides: $\int \frac{\sec^{2} y}{\tan y} \, dy = \int \frac{-e^{x}}{1 - e^{x}} \, dx$.
For the left side,let $\tan y = u$,then $\sec^{2} y \, dy = du$. Thus,$\int \frac{du}{u} = \ln|u| = \ln|\tan y|$.
For the right side,let $1 - e^{x} = t$,then $-e^{x} \, dx = dt$. Thus,$\int \frac{dt}{t} = \ln|t| = \ln|1 - e^{x}|$.
Combining these,we get: $\ln|\tan y| = \ln|1 - e^{x}| + \ln|C|$.
Using logarithmic properties: $\ln|\tan y| = \ln|C(1 - e^{x})|$.
Taking the exponential of both sides,we obtain the general solution: $\tan y = C(1 - e^{x})$.

(A) The given differential equation is:
$(x^{3}+x^{2}+x+1) \frac{dy}{dx} = 2x^{2}+x$
Factorizing the denominator: $(x^{2}(x+1) + 1(x+1)) \frac{dy}{dx} = 2x^{2}+x$
$(x+1)(x^{2}+1) \frac{dy}{dx} = 2x^{2}+x$
$dy = \frac{2x^{2}+x}{(x+1)(x^{2}+1)} dx$
Using partial fractions: $\frac{2x^{2}+x}{(x+1)(x^{2}+1)} = \frac{A}{x+1} + \frac{Bx+C}{x^{2}+1}$
$2x^{2}+x = A(x^{2}+1) + (Bx+C)(x+1)$
Setting $x = -1$: $2(-1)^{2} + (-1) = A((-1)^{2}+1) \Rightarrow 1 = 2A \Rightarrow A = \frac{1}{2}$
Comparing coefficients of $x^{2}$: $A+B = 2 \Rightarrow \frac{1}{2} + B = 2 \Rightarrow B = \frac{3}{2}$
Comparing constants: $A+C = 0 \Rightarrow \frac{1}{2} + C = 0 \Rightarrow C = -\frac{1}{2}$
Integrating: $\int dy = \int \frac{1/2}{x+1} dx + \int \frac{3/2x - 1/2}{x^{2}+1} dx$
$y = \frac{1}{2} \log|x+1| + \frac{3}{4} \log(x^{2}+1) - \frac{1}{2} \tan^{-1} x + C$
$y = \frac{1}{4} [2 \log|x+1| + 3 \log(x^{2}+1)] - \frac{1}{2} \tan^{-1} x + C$
$y = \frac{1}{4} \log[(x+1)^{2}(x^{2}+1)^{3}] - \frac{1}{2} \tan^{-1} x + C$
Given $y=1$ when $x=0$: $1 = \frac{1}{4} \log(1) - 0 + C \Rightarrow C = 1$
Thus,$y = \frac{1}{4} \log[(x+1)^{2}(x^{2}+1)^{3}] - \frac{1}{2} \tan^{-1} x + 1$

(C) $x(x^{2}-1) \frac{dy}{dx} = 1$
$\Rightarrow dy = \frac{dx}{x(x-1)(x+1)}$
Integrating both sides:
$\int dy = \int \frac{1}{x(x-1)(x+1)} dx \quad \dots(1)$
Using partial fractions:
$\frac{1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}$
$1 = A(x-1)(x+1) + Bx(x+1) + Cx(x-1)$
For $x=0, 1 = A(-1)(1) \Rightarrow A = -1$
For $x=1, 1 = B(1)(2) \Rightarrow B = \frac{1}{2}$
For $x=-1, 1 = C(-1)(-2) \Rightarrow C = \frac{1}{2}$
Substituting back into $(1)$:
$y = -\int \frac{1}{x} dx + \frac{1}{2} \int \frac{1}{x-1} dx + \frac{1}{2} \int \frac{1}{x+1} dx$
$y = -\log |x| + \frac{1}{2} \log |x-1| + \frac{1}{2} \log |x+1| + C_1$
$y = \frac{1}{2} \log \left| \frac{(x-1)(x+1)}{x^2} \right| + C_1 = \frac{1}{2} \log \left| \frac{x^2-1}{x^2} \right| + C_1$
Given $y=0$ when $x=2$:
$0 = \frac{1}{2} \log \left| \frac{4-1}{4} \right| + C_1 \Rightarrow 0 = \frac{1}{2} \log \left( \frac{3}{4} \right) + C_1 \Rightarrow C_1 = -\frac{1}{2} \log \left( \frac{3}{4} \right) = \frac{1}{2} \log \left( \frac{4}{3} \right)$
Thus,$y = \frac{1}{2} \log \left| \frac{x^2-1}{x^2} \right| + \frac{1}{2} \log \left( \frac{4}{3} \right) = \frac{1}{2} \log \left| \frac{4(x^2-1)}{3x^2} \right|$

(A) Given the differential equation: $\cos \left(\frac{dy}{dx}\right) = a$.
Taking $\cos^{-1}$ on both sides: $\frac{dy}{dx} = \cos^{-1}(a)$.
Integrating both sides with respect to $x$: $\int dy = \int \cos^{-1}(a) dx$.
This gives: $y = x \cos^{-1}(a) + C$ $(1)$.
Given the condition $y = 1$ when $x = 0$: $1 = 0 \cdot \cos^{-1}(a) + C$,which implies $C = 1$.
Substituting $C = 1$ into equation $(1)$,we get: $y = x \cos^{-1}(a) + 1$.

(A) Given differential equation is $\frac{dy}{dx} = y \tan x$.
Separating the variables,we get $\frac{dy}{y} = \tan x \, dx$.
Integrating both sides,$\int \frac{dy}{y} = \int \tan x \, dx$.
This gives $\ln |y| = \ln |\sec x| + C$,which can be written as $\ln |y| = \ln |C \sec x|$.
Thus,$y = C \sec x$ is the general solution.
Given that $y = 1$ when $x = 0$,we substitute these values: $1 = C \sec(0)$.
Since $\sec(0) = 1$,we have $1 = C \times 1$,so $C = 1$.
Substituting $C = 1$ into the general solution,we get the particular solution $y = \sec x$.

(A) The given differential equation is $xy \frac{dy}{dx} = (x+2)(y+2)$.
Separating the variables,we get:
$\left( \frac{y}{y+2} \right) dy = \left( \frac{x+2}{x} \right) dx$
$\Rightarrow \left( 1 - \frac{2}{y+2} \right) dy = \left( 1 + \frac{2}{x} \right) dx$
Integrating both sides:
$\int \left( 1 - \frac{2}{y+2} \right) dy = \int \left( 1 + \frac{2}{x} \right) dx$
$y - 2 \log |y+2| = x + 2 \log |x| + C$
$y - x - C = 2 \log |x| + 2 \log |y+2|$
$y - x - C = \log \left[ x^2 (y+2)^2 \right] \quad \dots(1)$
Since the curve passes through $(1, -1)$:
$-1 - 1 - C = \log \left[ (1)^2 (-1+2)^2 \right]$
$-2 - C = \log(1) = 0 \Rightarrow C = -2$
Substituting $C = -2$ into equation $(1)$:
$y - x + 2 = \log \left[ x^2 (y+2)^2 \right]$.

(A) Let the slope of the tangent at any point $(x, y)$ be $\frac{dy}{dx}$.
According to the problem,the product of the slope and the $y$-coordinate is equal to the $x$-coordinate:
$y \frac{dy}{dx} = x$
Separating the variables,we get:
$y \, dy = x \, dx$
Integrating both sides:
$\int y \, dy = \int x \, dx$
$\frac{y^{2}}{2} = \frac{x^{2}}{2} + C$
$y^{2} - x^{2} = 2C$
Let $2C = K$,so $y^{2} - x^{2} = K$.
The curve passes through $(0, -2)$,so substitute $x = 0$ and $y = -2$:
$(-2)^{2} - (0)^{2} = K$
$4 = K$
Thus,the equation of the curve is $y^{2} - x^{2} = 4$.

(C) Given the differential equation: $\frac{dy}{dx} = e^{x+y}$.
Using the property of exponents,we can write: $\frac{dy}{dx} = e^x \cdot e^y$.
Separating the variables,we get: $\frac{dy}{e^y} = e^x dx$,which is equivalent to $e^{-y} dy = e^x dx$.
Integrating both sides: $\int e^{-y} dy = \int e^x dx$.
This yields: $-e^{-y} = e^x + k$,where $k$ is the constant of integration.
Rearranging the terms: $e^x + e^{-y} = -k$.
Letting $C = -k$,we get the general solution: $e^x + e^{-y} = C$.
Thus,the correct option is $C$.

(A) Given the differential equation: $\left(x+y \frac{dy}{dx}\right)=1$.
Rearranging the terms,we get: $y \frac{dy}{dx} = 1 - x$.
This can be written as: $\frac{dy}{dx} = \frac{1-x}{y}$.
Separating the variables: $y \, dy = (1-x) \, dx$.
Integrating both sides: $\int y \, dy = \int (1-x) \, dx$.
$\frac{y^2}{2} = x - \frac{x^2}{2} + C_1$.
Multiplying by $2$: $y^2 = 2x - x^2 + 2C_1$.
Rearranging: $x^2 + y^2 - 2x = C$ (where $C = 2C_1$).
Thus,the general solution is $x^2 + y^2 - 2x = C$.

(N/A) The given differential equation is $\log \left(\frac{d y}{d x}\right)=3 x+4 y$.
By the definition of logarithm,we can write this as $\frac{d y}{d x}=e^{3 x+4 y}$.
This can be expressed as $\frac{d y}{d x}=e^{3 x} \cdot e^{4 y}$.
Separating the variables,we get $\frac{d y}{e^{4 y}}=e^{3 x} d x$,which is $e^{-4 y} d y=e^{3 x} d x$.
Integrating both sides,we get $\int e^{-4 y} d y=\int e^{3 x} d x$.
This gives $\frac{e^{-4 y}}{-4}=\frac{e^{3 x}}{3}+C$.
Multiplying by $12$,we get $-3 e^{-4 y}=4 e^{3 x}+12 C$,or $4 e^{3 x}+3 e^{-4 y}+K=0$,where $K=12 C$.
Given that $y=0$ when $x=0$,substituting these values into the equation: $4 e^{0}+3 e^{0}+K=0 \implies 4+3+K=0 \implies K=-7$.
Thus,the particular solution is $4 e^{3 x}+3 e^{-4 y}-7=0$.

(A) Given differential equation is $\frac{dy}{dx} + \sqrt{\frac{1-y^2}{1-x^2}} = 0$.
Rearranging the terms,we get $\frac{dy}{dx} = -\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$.
Separating the variables,we have $\frac{dy}{\sqrt{1-y^2}} = -\frac{dx}{\sqrt{1-x^2}}$.
Integrating both sides,we get $\int \frac{dy}{\sqrt{1-y^2}} = -\int \frac{dx}{\sqrt{1-x^2}}$.
This yields $\sin^{-1} y = -\sin^{-1} x + C$.
Therefore,the general solution is $\sin^{-1} x + \sin^{-1} y = C$.

Given differential equation: $\frac{dy}{dx} + \frac{y^{2}+y+1}{x^{2}+x+1} = 0$
Separating the variables,we get: $\frac{dy}{y^{2}+y+1} = -\frac{dx}{x^{2}+x+1}$
$\Rightarrow \frac{dy}{y^{2}+y+1} + \frac{dx}{x^{2}+x+1} = 0$
Integrating both sides: $\int \frac{dy}{(y+\frac{1}{2})^{2} + (\frac{\sqrt{3}}{2})^{2}} + \int \frac{dx}{(x+\frac{1}{2})^{2} + (\frac{\sqrt{3}}{2})^{2}} = C$
Using the formula $\int \frac{du}{u^{2}+a^{2}} = \frac{1}{a} \tan^{-1}(\frac{u}{a}) + C$,we get:
$\frac{2}{\sqrt{3}} \tan^{-1}(\frac{2y+1}{\sqrt{3}}) + \frac{2}{\sqrt{3}} \tan^{-1}(\frac{2x+1}{\sqrt{3}}) = C$
$\Rightarrow \tan^{-1}(\frac{2y+1}{\sqrt{3}}) + \tan^{-1}(\frac{2x+1}{\sqrt{3}}) = \frac{\sqrt{3}C}{2} = K$ (where $K$ is a constant)
Taking $\tan$ on both sides and using $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$:
$\frac{\frac{2y+1}{\sqrt{3}} + \frac{2x+1}{\sqrt{3}}}{1 - (\frac{2y+1}{\sqrt{3}})(\frac{2x+1}{\sqrt{3}})} = \tan K = A'$
$\Rightarrow \frac{2x+2y+2}{\sqrt{3}} = A'(1 - \frac{4xy+2x+2y+1}{3})$
$\Rightarrow \frac{2(x+y+1)}{\sqrt{3}} = A'(\frac{3-4xy-2x-2y-1}{3}) = A'(\frac{2-2x-2y-4xy}{3})$
$\Rightarrow x+y+1 = A(1-x-y-2xy)$,where $A$ is a new constant.